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AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.4

10th Class Maths 8th Lesson Similar Triangles Ex 8.4 Textbook Questions and Answers

Question 1.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Answer:

Given : □ ABCD is a rhombus.
Let its diagonals AC and BD bisect each other at ‘O’.
We know that “the diagonals in a rhombus are perpendicular to each other”.
In △AOD; AD² = OA² + OD² ………. (1)
[Pythagoras theorem]
In △COD; CD² = OC² + OD² ………. (2)
[Pythagoras theorem]
In △AOB; AB² = OA² + OB² ………. (3)
[Pythagoras theorem]
In △BOC; BC² = OB² + OC² ………. (4)
[Pythagoras theorem]
Adding the above equations we get AD² + CD² + AB² + BC² = 2 (OA² + OB2 + OC² + OD²)

Question 2.
ABC is a right triangle right angled at B. Let D and E be any points on AB and BC respectively. Prove that AE² + CD² = AC² + DE².

Answer:
Given: In △ABC; ∠B = 90°
D and E are points on AB and BC.
R.T.P.: AE² + CD² = AC² + DE²
Proof: In △BCD, △BCD is a right triangle right angled at B.
∴ BD² + BC² = CD² ……… (1)
[∵ Pythagoras theorem states that hypotenuse² = side² + side²]
In △ABE; ∠B = 90°
Adding (1) and (2), we get
BD² + BC² + AB² + BE² – CD² + AE²
(BD² + BE²) + (AB² + BC²) = CD² + AE²
DE² + AC² – CD² + AE² [Q.E.D.]
[∵ (i) In △DBE, ∠B = 90° and DE² = BD² + BE²
(ii) In △ABC, ∠B = 90° and AB² + BC²]

Question 3.
Prove that three times the square of any side of an equilateral triangle is equal to four times the square of the altitude.
Answer:
Given: △ABC, an equilateral triangle;
AD – altitude and the side is a units, altitude h units.
R.T.P: 3a² = 4h²

Proof: In △ABD, △ACD
∠B = ∠C [∵ 60°]
∠ADB = ∠ADC [∵ 90°]
∴ ∠BAD = ∠DAC [∵ Angle sum property]
Also, BA = CA
∴ △ABD s △ACD (by SAS congruence condition)
Hence, BD = CD = \(\frac{1}{2}\)BC = \(\frac{a}{2}\) [∵ c.p.c.t]
Now in △ABD, AB² = AD² + BD²
[∵ Pythagoras theorem]
a² = h² + \(\left(\frac{a}{2}\right)^{2}\)
a² = h² + \(\frac{a^{2}}{4}\)
h² = \(\frac{4 a^{2}-a^{2}}{4}\)
∴ h² = \(\frac{3 a^{2}}{4}\)
⇒ 4h² = 3a² (Q.E.D)

Question 4.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM . MR.
Answer:

Given: In △PQR, ∠P = 90° and PM ⊥ QR.
R.T.P : PM² = QM . MR
Proof: In △PQR; △MPR
∠P = ∠M [each 90°]
∠R = ∠R [common]
∴ △PQR ~ △MPR ……… (1)
[A.A. similarity]
In △PQR and △MQP,
∠P = ∠M (each 90°)
∠Q = ∠Q (common)
∴ △PQR ~ △MQP ……… (2)
[A.A. similarity]
From (1) and (2),
△PQR ~ △MPR ~ △MQP [transitive property]
∴ △MPR ~ △MQP
\(\frac{MP}{MQ}\) = \(\frac{PR}{QP}\) = \(\frac{MR}{MP}\)
[Ratio of corresponding sides of similar triangles are equal]
\(\frac{PM}{QM}\) = \(\frac{MR}{PM}\)
PM . PM = MR . QM
PM² = QM . MR [Q.E.D]

Question 5.
ABD is a triangle right angled at A and AC ⊥ BD.
Show that (i) AB² = BC BD
(ii) AD² = BD CD
(iii) AC² = BC DC.

Answer:
Given: In △ABD; ∠A = 90° AC ⊥ BD
R.T.P.:
i) AB² = BC . BD
Proof: In △ABD and △CAB,
∠BAD = ∠ACB [each 90°]
∠B = ∠B [common]
∴ △ABD ~ △CBA
[by A.A. similarity condition]
Hence, \(\frac{AB}{BC}\) = \(\frac{BD}{AB}\) = \(\frac{AD}{AC}\)
[∵ Ratios of corresponding sides of similar triangles are equal]
\(\frac{AB}{BD}\) = \(\frac{BC}{AB}\)
⇒ AB . AB = BC . BD
∴ AB² = BC . BD

ii) AD² = BD . CD
Proof: In △ABD and △CAD
∠BAD = ∠ACD [each 90°]
∠D = ∠D (common)
∴ △ABD ~ △CAD [A.A similarity]
Hence, \(\frac{AB}{AC}\) = \(\frac{BD}{AD}\) = \(\frac{AD}{CD}\)
⇒ \(\frac{BD}{AD}\) = \(\frac{AD}{CD}\)
AD . AD = BD . CD
AD² = BD . CD [Q.E.D]

iii) AC² = BC . DC
Proof: From (i) and (ii)
△ACB ~ △DCA
[∵ △BAD ~ △BCA ~ △ACD
Hence, \(\frac{AC}{DC}\) = \(\frac{BC}{AC}\) = \(\frac{AB}{AD}\)
\(\frac{AC}{DC}\) = \(\frac{BC}{AC}\)
AC . AC = BC . DC
AC² = BC . DC [Q.E.D]

Question 6.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Answer:

Given: In △ABC; ∠C = 90°; AC = BC.
R.T.P.: AB² = 2AC²
Proof: In △ACB; ∠C = 90°
Hence, AC² + BC² = AB²
[Square of the hypotenuse is equal to sum of the squares of the other two sides – Pythagoras theorem]
⇒ AC² + AC² = AB² [∵ AC = BC given]
⇒ AB² = 2AC² [Q.E.D.]

Question 7.
‘O’ is any point in the interior of a triangle ABC.
OD ⊥ BC, OE ⊥ AC and OF ⊥ AB, show that
i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
ii) AF² + BD² + CE² = AE² + CD² + BF².

Answer:
Given: △ABC; O’ is an interior point of △ABC.
OD ⊥ BC, OE ⊥ AC, OF ⊥ AB.
R.T.P.:
i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
Proof: In OAF, OA² = AF² + OF² [Pythagoras theorem]
⇒ OA² – OF² = AF² …….. (1)
In △OBD,
OB² = BD² + OD²
⇒ OB² – OD² = BD² …….. (2)
In △OCE, OC² = CE² + OE²
OC² – OE² = CE² ……… (3)
Adding (1), (2) and (3) we get,
OA² – OF² + OB² – OD² + OC² – OE² = AF² + BD² + CE²
OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE² ……… (4)

ii) AF² + BD² + CE² = AE² + CD² + BF²
In △OAE,
OA² = AE² + OF² ……… (1)
⇒ OA² – OE² = AE²
In △OBF, OB² = BF² + OF²
OB² – OF² = BF² ……… (2)
In △OCD, OC² = OD² + CD²
OC² – OD² = CD² ……… (3)
Adding (1), (2) and (3) we get
OA² – OE² + OB² – OF² + OC² – OD² = AE² + BF² + CD²
⇒ OA² + OB² + OC² – OD² – OE² – OF² = AE² + CD² + BF²
⇒ AF² + BD² + CE² = AE² + CD² + BF² [From problem (i)]

Question 8.
A wire attached to vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Answer:

Height of the pole AB = 18 m.
Length of the wire AC = 24 m.
Distance beween the pole and the stake be ‘d’ meters.
By Pythagoras theorem,
Hypotenuse² = side² + side²
24² = 18² + d²
d² = 24² – 18² = 576 – 324 = 252
= \(\sqrt{36 \times 7}\)
∴ d = 6√7 m.

Question 9.
Two poles of heights 6 m and 11m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Answer:

Let the height of the first pole AB = 6 m.
Let the height of the second pole CD = 11 m.
Distance between the poles AC = 12 m.
From the figure □ ACEB is a rectangle.
∴ AB = CE = 6 m
ED = CD – CE = 11 – 6 = 5 m
Now in △BED; ∠E = 90°; DE = 5 m; BE = 12 m
BD² = BE² + DE²
[hypotenuse² = side² + side² – Pythagoras theorem]
= 12² + 5²
= 144 + 25
BD² = 169
BD = √l69 = 13m
∴ Distance between the tops of the poles = 13 m.

Question 10.
In an equilateral triangle ABC, D is a point on side BC such that BD = \(\frac{1}{3}\) BC. Prove that 9AD² = 7AB².
Answer:

In △ABE, ∠E = 90°
⇒ \(\overline{\mathrm{AB}}\) is hypotenuse.
∴ AB² = AE² + BE²
AE² = AB² – BE²
⇒ AE² = AB² – \(\left(\frac{BC}{2}\right)^{2}\)
= AE² = AB² – \(\left(\frac{AB}{2}\right)^{2}\) (∵ AB = BC)
⇒ AE² = \(\frac{3}{4}\)AB² ……… (1)
In △ADE, ∠E = 90°
⇒ \(\overline{\mathrm{AD}}\) is hypotenuse.
⇒ AD² = AE² + DE²
⇒ AE² = AD² + DE²

⇒ 28 AB² = 36 AD²
⇒ 7 AB² = 9 AD²
Hence proved.

Question 11.
In the given figure, ABC is a triangle right angled at B. D and E are points on BC trisect it. Prove that 8 AE² = 3 AC² + 5 AD².

Answer:
In △ABC, ∠B=90°
⇒ \(\overline{\mathrm{AD}}\) is hypotenuse.
AC2 = AB2 + BC2
3AC² = 3AB² + 3BC² …….. (1)
In △ABD, ∠B = 90°
⇒ AD is hypotenuse.
∴ AD² = AB² + BD² = AB² + \(\left(\frac{BC}{3}\right)^{2}\)
⇒ AD² = AB² + \(\frac{\mathrm{BC}^{2}}{9}\)
⇒ 5 AD² = 5 AB² + \(\frac{5 \mathrm{BC}^{2}}{9}\) …….. (2)
(1) + (2)
3 AC² + 5 AD² = 3 AB² + 3 BC² + 5 AB² + \(\frac{5}{9} \mathrm{BC}^{2}\)
= 8AB² + \(\frac{32}{9} \mathrm{BC}^{2}\) ……… (3)
Now in △ABE, ∠B = 90°
⇒ \(\overline{\mathrm{AE}}\) is hypotenuse.
⇒ AE² = AB² + BE² = AB² + \(\left(\frac{2}{3} BC\right)^{2}\)
= AB² + \(\frac{4}{9} \mathrm{BC}^{2}\)
⇒ AE² = 8AB² + \(\frac{32}{9} \mathrm{BC}^{2}\) ……… (4)
∴ RHS of (3) and (4) are equal.
∴ LHS of (3) and (4) are equal.
∴ 8 AE² = 3 AC² + 5 AD².
Hence proved.

Question 12.
ABC is an isosceles triangle right angled at B. Equilateral triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of △ABE and △ACD.
Answer:

Given: △ABC, AB = BC and ∠B = 90°
△ABE on AB; △ACD on AC are equiangular triangles.
Let equal sides of the isosceles right triangle, AB = BC = a (say)
Then, in △ABC, ∠B = 90°
AC² – AB² + BC²
[hypotenuse² = side² + side² – Pythagoras theorem] = a² + a² = 2a²
Since, △ABE ~ △ACD
\(\frac{\Delta \mathrm{ABE}}{\Delta \mathrm{ACD}}\) = \(\frac{\mathrm{AB}^{2}}{\mathrm{AC}^{2}}\)
[∵ Ratio of areas of two similar tri-angles is equal to the ratio of squares of their corresponding sides]
= \(\frac{a^{2}}{2 a^{2}}\) = \(\frac{1}{2}\)
△ABE : △ACD = 1 : 2.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.3

10th Class Maths 8th Lesson Similar Triangles Ex 8.3 Textbook Questions and Answers

Question 1.
Equilateral triangles are drawn on the three sides of a right angled triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.
Answer:

Let △PQR is a right angled triangle, ∠Q = 90°
Let PQ = a, QR b and
PR = hypotenuse = c
Then from Pythagoras theorem we can
say a² + b² = c² ……… (1)
△PSR is an equilateral triangle drawn on hypotenuse
∴ PR = PS = RS = c,
Then area of triangle on hypotenuse
= \(\frac{\sqrt{3}}{4}\)c² ……… (2)
△QRU is an equilateral triangle drawn on the side ‘QR’ = b
∴ QR = RU = QU = b
Then area of equilateral triangle drawn on the side = \(\frac{\sqrt{3}}{4}\)b² …….. (3)
△PQT is an equilateral triangle drawn on another side ‘PQ’ = a
∴ PQ = PT = QT = a
Area of an equilateral triangle drawn an another side ‘PQ’ = \(\frac{\sqrt{3}}{4}\)a² …….. (4)
Now sum of areas of equilateral triangles on the other two sides

= Area of equilateral triangle on the hypotenuse.
Hence Proved.

Question 2.
Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangles described on its diagonal.
Answer:

Let PQRS is square whose side is ‘a’ units then PQ = QR = RS = SP = ‘a’ units.
Then the diagonal
\(\overline{\mathrm{PR}}\) = \(\sqrt{a^{2}+a^{2}}\) = a√2 units.
Let △PRT is an equilateral triangle, then PR = RT = PT = a√2
∴ Area of equilateral triangle constructed on diagonal

Let △QRZ is another equilateral triangle whose sides are
\(\overline{\mathrm{QR}}\) = \(\overline{\mathrm{RZ}}\) = \(\overline{\mathrm{QZ}}\) = ‘a’ units
Then the area of equilateral triangle constructed on one side of square = \(\frac{\sqrt{3}}{4}\)a² ……. (2)
∴ \(\frac{1}{2}\) of area of equilateral triangle on diagonal = \(\frac{1}{2}\left(\frac{\sqrt{3}}{2} a^{2}\right)\) = \(\frac{\sqrt{3}}{4}\)a² = area of equilateral triangle on the side of square.
Hence Proved.

Question 3.
D, E, F are midpoints of sides BC, CA, AB of △ABC. Find the ratio of areas of △DEF and △ABC.
Answer:

Given in △ABC D, E, F are the midpoints of the sides BC, CA and AB.
In △ABC, EF is the line join of mid-points of two sides AB and AC of △ABC.
Thus FE || BC [∵ \(\frac{AF}{FB}\) = \(\frac{AE}{EC}\) Converse of B.P.T.]
Similarly DE divides AC and BC in the same ratio, i.e., DE || AB.
Now in □ BDEF, both pairs of opposite sides (BD || EF and DE || BF) are parallel.
Hence □ BDEF is a parallelogram where DF is a diagonal.
∴ △BDF ≃ △DEF ……… (1)
Similarly we can prove that
△DEF ≃ △CDE ……… (2)
[∵ CDFE is a parallelogram]
Also, △DEF ≃ △AEF …….. (3)
[∵ □ AEDF is a parallelogram]
From (1), (2) and (3)
△AEF ≃ △DEF ≃ BDF ≃ △CDE
Also, △ABC = △AEF + △DEF + △BDF + △CDE = 4 . △DEF
Hence, △ABC : △DEF = 4 : 1.

Question 4.
In △ABC, XY || AC and XY divides the triangle into two parts of equal area. Find the ratio of \(\frac{AX}{XB}\).
Answer:

Given: In △ABC, XY || AC.
XY divides △ABC into two points of equal area.
In △ABC, △XBY
∠B = ∠B
∠A = ∠X
∠C = ∠Y
[∵ XY || AC; (∠A, ∠X) and ∠C, ∠Y are the pairs of corresponding angles]
Thus △ABC ~ △XBY by A.A.A similarity condition.
Hence \(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{XBY}}=\frac{\mathrm{AB}^{2}}{\mathrm{XB}^{2}}\)
[∵ The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides]

⇒ \(\frac{AX}{XB}\) + 1 = √2
⇒ \(\frac{AX}{XB}\) = √2 – 1
Hence the ratio \(\frac{AX}{XB}\) = \(\frac{√2 – 1}{1}\).

Question 5.
Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Answer:

Given: △ABC ~ △XYZ
R.T.P: \(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{XYZ}}=\frac{\mathrm{AD}^{2}}{\mathrm{XW}^{2}}\)
Proof : We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Hence the ratio of areas of two similar triangles is equal to the squares of ratio of their corresponding medians.

Question 6.
△ABC ~ △DEF. BC = 3 cm, EF = 4 cm and area of △ABC = 54 cm². Determine the area of △DEF.
Answer:

Given: △ABC ~ △DEF
BC = 3 cm
EF = 4 cm
△ABC = 54 cm²
∴ △ABC ~ △DEF, we have
\(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{DEF}}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}\)
[∵ The ratio of two similar triangles is equal to the ratio of the squares of the corresponding sides].
\(\frac{54}{\Delta \mathrm{DEF}}\) = \(\frac{3^{2}}{4^{2}}\)
∴ △DEF = \(\frac{54 \times 16}{9}\) = 96 cm²

Question 7.
ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm and BP = 3 cm, AQ =1.5 cm, CQ = 4.5 cm. Prove that area of △APQ = \(\frac{1}{16}\) (area of △ABC).
Answer:

Given: △ABC and \(\overline{\mathrm{PQ}}\) – a line segment meeting AB in P and AC in Q.
AP = 1 cm; AQ =1.5 cm;
BP = 3 cm; CQ = 4.5 cm.
\(\frac{AP}{PQ}\) = \(\frac{1}{3}\) ……… (1)
\(\frac{AQ}{QC}\) = \(\frac{1.5}{4.5}\) = \(\frac{1}{3}\) ……..(2)
From (1) and (2),
\(\frac{AP}{BP}\) = \(\frac{AQ}{CQ}\)
[i.e., PQ divides AB and AC in the same ratio – By converse of Basic pro-portionality theorem]
Hence, PQ || BC.
Now in △APQ and △ABC
∠A = ∠A (Common)
∠P = ∠B [∵ Corresponding angles for the parallel lines PQ and BC]
∠Q = ∠C
∴ △APQ ~ △ABC [∵ A.A.A similarity condition]
Now, \(\frac{\Delta \mathrm{APQ}}{\Delta \mathrm{ABC}}=\frac{\mathrm{AP}^{2}}{\mathrm{AB}^{2}}\)
[∵ Ratio of two similar triangles is equal to the ratio of the squares of their corresponding sides].
= \(\frac{1^{2}}{(3+1)^{2}}\) = \(\frac{1}{16}\) [∵ AB = AP + BP = 1 + 3 = 4 cm]
∴ △APQ = \(\frac{1}{16}\) (area of △ABC) [Q.E.D]

Question 8.
The areas of two similar triangles are 81 cm² and 49 cm² respectively. If the altitude of the bigger triangle is 4.5 cm. Find the corresponding altitude of the smaller triangle.

Answer:
Given: △ABC ~ △DEF
△ABC = 81 cm²
△DEF = 49 cm²
AX = 4.5 cm
To find: DY
We know that,
\(\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{DEF}}=\frac{\mathrm{AX}^{2}}{\mathrm{DY}^{2}}\)
[∵ Ratio of areas of two similar triangles is equal to ratio of the squares of their corresponding altitudes]

∴ DY = 3.5 cm.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.2

10th Class Maths 8th Lesson Similar Triangles Ex 8.2 Textbook Questions and Answers

Question 1.
In the given figure, ∠ADE = ∠B
i) Show that △ABC ~ △ADE
ii) If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2 cm, find DE.

Answer:
i) Given: △ABC and ∠ADE = ∠B
R.T.P: △ABC ~ △ADE.
Proof: In △ABC and △ADE
∠A = ∠A [∵ Common]
∠B = ∠ADE [∵ Given]
∴ ∠C = ∠AED [∵ By Angle Sum property of triangles] △ABC ~ △ADE by AAA similarity condition.]

ii) AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2 cm, find DE.
To find DE; △ABC ~ △ADE
Hence,
\(\frac{AB}{AD}\) = \(\frac{BC}{DE}\) = \(\frac{AC}{AE}\)
[∵ Ratios of corresponding sides are equal]

Question 2.
The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of the first triangle is 12 cm, determine the corresponding side of the second triangle.
Answer:

Given: △ABC ~ △PQR
Perimeter of △ABC = 30 cm.
Perimeter of △PQR = 20 cm.
AB = 12 cm.
To find: \(\overline{\mathrm{PQ}}\)
Ratio of perimeters = 30 : 20 = 3 : 2
Let the length of the side corresponding to the side with length 12 cm be x.
Then 30 : 20 : : 12 : x
30x = 20 x 12
\(x = \frac{20 \times 12}{30}\) = 8 cm

Question 3.
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp-post is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Answer:

Given:
A lamp-post \(\overline{\mathrm{AB}}\) of height = 3.6 m
= 360 cm.
Speed of the girl = 1.2 m/sec.
Distance travelled in 4 sec = Speed x Time = 1.2 × 4 = 4.8 m = 480 cm.
\(\overline{\mathrm{CD}}\), height of the girl = 90 cm.
Let the length of the shadow at a distance of 4.8 m from the lamp post = x cm.
From the figure,
△ABE ~ △DCE
[∵ ∠B = ∠C = 90°
∠E = ∠C common
(A.A. similarity)]
Hence,
\(\frac{AB}{DC}\) = \(\frac{BE}{CE}\) = \(\frac{AE}{DE}\)
∴ \(\frac{360}{90}\) = \(\frac{480+x}{x}\)
⇒ 4 = \(\frac{480+x}{x}\)
⇒ 4x = 480 + x
⇒ 4x – x = 480
⇒ 3x = 480
⇒ x = 160 cm = 1.6 m
∴ Length of the shadow = 1.6 m

Question 4.
CM and RN are respectively the medians of similar triangles △ABC and △PQR. Prove that
i) △AMC ~ △PNR
ii) \(\frac{CM}{RN}\) = \(\frac{AB}{PQ}\)
iii) △CMB ~ △RNQ

Answer:
Given : △ABC ~ △PQR
CM is a median through C of △ABC.
RN is a median through R of △PQR.
R.T.P:
i) △AMC ~ △PNR.
Proof: In △AMC and △PNR,
\(\frac{AC}{PR}\) = \(\frac{AM}{PN}\) and ∠A = ∠P [∵ In △ABC, △PQR and M, N are the mid-points of AB and PQ]
∴ △AMC ~ △PNR
[∵ SAS similarity condition]

ii) \(\frac{CM}{RN}\) = \(\frac{AB}{PQ}\)
Proof: From (i) we have
△AMC ~ △PNR
Hence \(\frac{AC}{PQ}\) = \(\frac{AM}{PN}\) = \(\frac{CM}{RN}\)
[∵ Ratio of corresponding sides of two similar triangles are equal]
Thus, \(\frac{CM}{RN}\) = \(\frac{AM×2}{PN×2}\)
[Multiplying both numerator and the denominator by 2]
\(\frac{CM}{RN}\) = \(\frac{AB}{PQ}\) [2AM = AB; 2PN = PQ]

iii) △CMB ~ △RNQ
Proof: In △CMB and △RNQ
∠B = ∠Q [Corresponding angles of △ABC and △PQR]
Also, \(\frac{BC}{RQ}\) = \(\frac{BM}{QN}\)

Thus, △CMB ~ △RNQ by S.A.S similarity condition.

Question 5.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point ‘O’. Using the criterion of similarity for two triangles, show that \(\frac{OA}{OC}\) = \(\frac{OB}{OD}\).
Answer:

Given : □ ABCD, AB || DC
The diagonals AC and BD intersect at ‘O’.
R.T.P: \(\frac{OA}{OC}\) = \(\frac{OB}{OD}\)
Construction: Draw EF || AB, passing through ‘O’.
Proof: In △ACD, OE || CD [∵ Construction]
Hence \(\frac{OA}{OC}\) = \(\frac{EA}{ED}\) …….. (1)
(∵ Line drawn parallel to one side of a triangle divides other two sides in the same ratio – Basic proportionality theorem)
Also in △ABD, EO || AB [Construction] Hence,
\(\frac{EA}{ED}\) = \(\frac{OB}{OD}\) ……… (2)
(∵ Basic proportionality theorem) From (1) and (2), we have
\(\frac{OA}{OC}\) = \(\frac{OB}{OD}\)
∴ Hence proved.

Question 6.
AB, CD, PQ are perpendicular to BD. AB = x, CD = y and PQ = z, prove that \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\).

Answer:
Given ∠B = ∠Q = ∠D = 90°
Thus, AB || PQ || CD.
Now in △BQP, △BDC
∠B = ∠B (Common)
∠Q = ∠D (90°)
∠P = ∠C [∵ Angle Sum property of triangles]
∴ △BQP ~ △BDC
(by A.A.A similarity condition)
Hence \(\frac{BQ}{BD}\) = \(\frac{PQ}{CD}\)
[∵ Ratio of corresponding sides is equal] Also in △DQP and △DBA
∠D = ∠D (Common)
∠Q = ∠B (90°)
∴ △DQP ~ △DBA (by A.A. similarity condition)
\(\frac{QD}{BD}\) = \(\frac{PQ}{AB}\)
[ Ratio of corresponding sides is equal]
Adding (1) and (2), we get

Question 7.
A flag pole 4 m tall casts a 6 m., shadow. At the same time, a nearby building casts a shadow of 24 m. How tall is the building?
Answer:
Given: 4 m length flag pole casts a shadow 6 m.
Let x m length/tall building casts a shadow 24 m.

Let AB be the length of flag pole = 4 m.
Shadow of AB = BC = 6 m.
PQ be the building = x m (say)
QR, the shadow of the building = 24 m
From the figure,
∠A = ∠P
∠B = ∠Q
∴ △ABC ~ △PQR by A.A. similarity condition
Hence \(\frac{AB}{PQ}\) = \(\frac{BC}{QR}\)
[∵ Ratio of corresponding angles is equal]
\(\frac{4}{6}\) = \(\frac{x}{24}\)
x = \(\frac{24 \times 4}{6}\) = 16 m
∴ Height of the building = 16 m.

Question 8.
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of △ABC and △FEG respectively. If △ABC ~ △FEG then show that
i) \(\frac{CD}{GH}\) = \(\frac{AC}{FG}\)
ii) △DCB ~ △HGE
iii) △DCA ~ △HGF
Answer:

Given: △ABC ~ △FEG.
CD is the bisector of ∠C and GH is the bisector of ∠G.
R.T.P.:
i) \(\frac{CD}{GH}\) = \(\frac{AC}{FG}\)
In △ACD and △FGH
∠A = ∠F
[∵ Corresponding angles of △ABC and △FEG]
∠ACD = ∠FGH
[∵ ∠C = ∠G ⇒ \(\frac{1}{2}\)∠C = \(\frac{1}{2}\)∠G ⇒ ∠ACD = ∠FGH]
∴ By A.A. similarity condition, △ACD ~ △FGH
\(\frac{AC}{FG}\) = \(\frac{CD}{GH}\) = \(\frac{AD}{FH}\)
[∵ Ratio of the Corresponding angles is equal]
⇒ \(\frac{AC}{FG}\) = \(\frac{CD}{GH}\) [Q.E.D]

ii) △DCB ~ △HGE
In △DCB and △HGE,
∠B = ∠E
[∵ Corresponding angles of △ABC and △FEG]
∠DCB = ∠HGE
[∵ ∠C = ∠G ⇒ \(\frac{1}{2}\)∠C = \(\frac{1}{2}\)∠G ⇒ ∠DCB = ∠HGE]
∴ △DCB ~ △HGE . (by A.A. similarity condition)

iii) △DCA ~ △HGF
In △DCA and △HGF
∠A = ∠F
\(\frac{1}{2}\)∠C = \(\frac{1}{2}\)∠G ⇒ ∠DCA = ∠HGF
[∵ Corresponding angles of the similar triangles]
∴ △DCA ~ △HGF
[ A.A. similarity condition]

Question 9.
AX and DY are altitudes of two similar triangles △ABC and △DEF. Prove that AX : DY = AB : DE.
Answer:

Given: △ABC ~ △DEF.
AX ⊥ BC and DY ⊥ EF.
R.T.P.: AX : DY = AB : DE.
Proof: In △ABX and △DEY ∠B = ∠E [∵ Corresponding angles of △ABC and △DEF]
∠AXB = ∠DYE [given]
∴ △ABX ~ △DEY
(by A.A. similarity condition)
Hence \(\frac{AB}{DE}\) = \(\frac{BX}{EY}\) = \(\frac{AX}{DY}\)
[∵ Ratios of corresponding sides of similar triangles are equal]
⇒ AX : DY = AB : DE [Q.E.D.]

Question 10.
Construct a triangle shadow similar to the given △ABC, with its sides equal to \(\frac{5}{3}\) of the corresponding sides of the triangle ABC.
Answer:

Steps of construction :

1. Draw a △ABC with certain measures.
2. Draw a ray \(\overrightarrow{\mathrm{BX}}\) making an acute angle with BC on the side opposite to vertex A.
3. Locate 8 points (B₁, B₂, …., B₈) on \(\overrightarrow{\mathrm{BX}}\) such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇ = B₇B₈.
4. Join B₅, C.
5. Draw a line parallel to B₅C through which it intersects BC extended at C’.
6. Draw a line parallel to AC through ‘C’ which meets \(\overrightarrow{\mathrm{BA}}\) produced at A’.
7. △A’BC’ is the required triangle.

Question 11.
Construct a triangle of sides 4 cm, 5 cm and 6 cm. Then, construct a triangle similar to it,whose sides are 2/3 of the corresponding sides of the first triangle.
Answer:

Steps of construction:

1. Draw △ABC with AB = 4 cm, BC = 5 cm and CA = 6 cm.
2. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
3. Mark three points B₁, B₂ and B₃ on \(\overrightarrow{\mathrm{BX}}\) such that BB₁ = B₁B₂ = B₂B₃.
4. Join B₃, C.
5. Draw a line parallel to B₃C through B₂ meeting BC at C’.
6. Draw a line parallel to BA through C’ meeting BA at A’.
7. △A’BC’ is the required triangle.

Question 12.
Construct an isosceles triangle whose base is 8 cm and altitude is 4 cm. Then, draw another triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Answer:

Steps of construction:

1. Draw AABC in which BC = 8 cm and altitude AD = 4 cm.
2. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.
3. Mark three points B₁, B₂ and B₃ such that BB₁ = B₁B₂ = B₂B₃.
4. Join B₂C.
5. Draw a line parallel to B₂C through B₃ meeting BC produced C’.
6. Draw a line paral1e1 to AC through C’ meeting BA produced at A’.
7. △A’BC’ is the required triangle.

 

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.1

10th Class Maths 8th Lesson Similar Triangles Ex 8.1 Textbook Questions and Answers

Question 1.
In △PQR, ST is a line such that \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and also ∠PST = ∠PRQ.
Prove that △PQR is an isosceles triangle.
Answer:
Given : In △PQR,
\(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and ∠PST= ∠PRQ.

R.T.P: △PQR is isosceles.
Proof: \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\)
Hence, ST || QR (Converse of Basic proportionality theorem)
∠PST = ∠PQR …….. (1)
(Corresponding angles for the lines ST || QR)
Also, ∠PST = ∠PRQ ……… (2) given
From (1) and (2),
∠PQR = ∠PRQ
i.e., PR = PQ
[∵ In a triangle sides opposite to equal angles are equal]
Hence, APQR is an isosceles triangle.

Question 2.
In the given figure, LM || CM and LN || CD. Prove that \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).

Answer:
Given : LM || CB and LN || CD In △ABC, LM || BC (given) Hence,

Adding ‘1’ on both sides.

From (1) and (2)
∴ \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).

Question 3.
In the given figure, DE || AC and DF || AE. Prove that \(\frac{BF}{FE}\) = \(\frac{BE}{AC}\).

Answer:
In △ABC, DE || AC
Hence \(\frac{BE}{EC}\) = \(\frac{BD}{DA}\) ………. (1)
[∵ A line drawn parallel to one side of a triangle divides the other two sides in the same ratio – Basic proportionality theorem]
Also in △ABE, DF || AE
Hence \(\frac{BF}{FE}\) = \(\frac{BD}{DA}\) ………. (2)
From (1) and (2), \(\frac{BF}{FE}\) = \(\frac{BE}{AC}\) Hence proved.

Question 4.
Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (Using Basic proportionality theorem).
Answer:

Given: In △ABC; D is the mid-point of AB.
A line ‘l’ through D, parallel to BC, meeting AC at E.
R.T.P: E is the midpoint of AC.
Proof:
DE || BC (Given)
then
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)(From Basic Proportional theorem)
Also given ‘D’ is mid point of AB.
Then AD = DB.
⇒ \(\frac{AD}{DB}\) = \(\frac{DB}{DB}\) = \(\frac{AE}{EC}\) = 1
⇒ AE = EC
∴ ‘E’ is mid point of AC
∴ The line bisects the third side \(\overline{\mathrm{AC}}\).
Hence proved.

Question 5.
Prove that a line joining the mid points of any two sides of a triangle is parallel to the third side. (Using converse of Basic proportionality theorem)
Answer:
Given: △ABC, D is the midpoint of AB and E is the midpoint of AC.
R.T.P : DE || BC.
Proof:

Since D is the midpoint of AB, we have AD = DB ⇒ \(\frac{AD}{DB}\) = 1 ……. (1)
also ‘E’ is the midpoint of AC, we have AE = EC ⇒ \(\frac{AE}{EC}\) = 1 ……. (2)
From (1) and (2)
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)
If a line divides any two sides of a triangle in the same ratio then it is parallel to the third side.
∴ DE || BC by Basic proportionality theorem.
Hence proved.

Question 6.
In the given figure, DE || OQ and DF || OR. Show that EF || QR.

Answer:
Given: △PQR, DE || OQ; DF || OR
R.T.P: EF || QR
Proof:
In △POQ;
\(\frac{PE}{EQ}\) = \(\frac{PD}{DO}\) ……. (1)
[∵ ED || QO, Basic proportionality theorem]
In △POR; \(\frac{PF}{FR}\) = \(\frac{PD}{DO}\) ……. (2) [∵ DF || OR, Basic Proportionality Theorem]
From (1) and (2),
\(\frac{PE}{EQ}\) = \(\frac{PF}{FR}\)
Thus the line \(\overline{\mathrm{EF}}\) divides the two sides PQ and PR of △PQR in the same ratio.
Hence, EF || QR. [∵ Converse of Basic proportionality theorem]

Question 7.
In the given figure A, B and C are points on OP, OQ and OR respec¬tively such that AB || PQ and AC || PR. Show that BC || QR.

Answer:
Given:
In △PQR, AB || PQ; AC || PR
R.T.P : BC || QR
Proof: In △POQ; AB || PQ
\(\frac{OA}{AP}\) = \(\frac{OB}{BQ}\) ……… (1)
(∵ Basic Proportional theorem)
and in △OPR, Proof: In △POQ; AB || PQ
\(\frac{OA}{AP}\) = \(\frac{OC}{CR}\) ……… (2)
From (1) and (2), we can write
\(\frac{OB}{BQ}\) = \(\frac{OC}{CR}\)
Then consider above condition in △OQR then from (3) it is clear.
∴ BC || QR [∵ from converse of Basic Proportionality Theorem]
Hence proved.

Question 8.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point ‘O’. Show that\(\frac{AO}{BO}\) = \(\frac{CO}{DO}\).
Answer:
Given: In trapezium □ ABCD, AB || CD. Diagonals AC, BD intersect at O.
R.T.P: \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\)

Construction:
Draw a line EF passing through the point ‘O’ and parallel to CD and AB.
Proof: In △ACD, EO || CD
∴ \(\frac{AO}{CO}\) = \(\frac{AE}{DE}\) …….. (1)
[∵ line drawn parallel to one side of a triangle divides other two sides in the same ratio by Basic proportionality theorem]
In △ABD, EO || AB
Hence, \(\frac{DE}{AE}\) = \(\frac{DO}{BO}\)
[∵ Basic proportionality theorem]
\(\frac{BO}{DO}\) = \(\frac{AE}{ED}\) …….. (2) [∵ Invertendo]
From (1) and (2),
\(\frac{AO}{CO}\) = \(\frac{BO}{DO}\)
⇒ \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\) [∵ Alternendo]

Question 9.
Draw a line segment of length 7.2 cm and divide it in the ratio 5 : 3. Measure the two parts.
Answer:

Steps of construction:

1. Draw a line segment \(\overline{\mathrm{AB}}\) of length 7.2 cm.
2. Construct an acute angle ∠BAX at A.
3. Mark off 5 + 3 = 8 equal parts (A₁, A₂, …., A₈) on \(\stackrel{\leftrightarrow}{\mathrm{AX}}\) with same radius.
4. Join A₈ and B.
5. Draw a line parallel to \(\stackrel{\leftrightarrow}{\mathrm{A}_{8} \mathrm{~B}}\) at A₅ meeting AB at C.
6. Now the point C divides AB in the ratio 5:3.
7. Measure AC and BC. AC = 4.5 cm and BC = 2.7 cm.

 

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.3

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.3 Textbook Questions and Answers

Question 1.
A chord of a circle of radius 10 cm. subtends a right angle at the centre. Find the area of the corresponding: (use π = 3.14)
i) Minor segment ii) Major segment
Answer:

Angle subtended by the chord = 90° Radius of the circle = 10 cm
Area of the minor segment = Area of the sector POQ – Area of △POQ
Area of the sector = \(\frac{x}{360}\) × πr²
\(\frac{90}{360}\) × 3.14 × 10 × 10 = 78.5
Area of the triangle = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 10 × 10 = 50
∴ Area of the minor segment = 78.5 – 50 = 28.5 cm²
Area of the major segment = Area of the circle – Area of the minor segment
= 3.14 × 10 × 10 – 28.5
= 314 – 28.5 cm²
= 285.5 cm²

Question 2.
A chord of a circle of radius 12 cm. subtends an angle of 120° at the centre. Find the area of the corresponding minor segment of the circle.
(use π = 3.14 and √3 = 1.732)
Answer:
Radius of the circle r = 12 cm.
Area of the sector = \(\frac{x}{360}\) × πr²
Here, x = 120°

\(\frac{120}{360}\) × 3.14 × 12 × 12 = 150.72
Drop a perpendicular from ‘O’ to the chord PQ.
△OPM = △OQM [∵ OP = OQ ∠P = ∠Q; angles opp. to equal sides OP & OQ; ∠OMP = ∠OMQ by A.A.S]
∴ △OPQ = △OPM + △OQM = 2 . △OPM
Area of △OPM = \(\frac{1}{2}\) × PM × OM

= 18 × 1.732 = 31.176 cm
∴ △OPQ = 2 × 31.176 = 62.352 cm²
∴ Area of the minor segment
= (Area of the sector) – (Area of the △OPQ)
= 150.72 – 62.352 = 88.368 cm²

Question 3.
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm. sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. (use π = \(\frac{22}{7}\))
Answer:
Angle made by the each blade = 115°
Total area swept by two blades
= Area of the sector with radius 25 cm and angle 115°+ 115° = 230°
= Area of the sector = \(\frac{x}{360}\) × πr²
= \(\frac{230}{360}\) × \(\frac{22}{7}\) × 25 × 25
= 1254.96
≃  1255 cm²

Question 4.
Find the area of the shaded region in figure, where ABCD is a square of side 10 cm. and semicircles are drawn with each side of the square as diameter (use π = 3.14).

Answer:
Let us mark the four unshaded regions as I, II, III and IV.

Area of I + Area of II
= Area of ABCD – Areas of two semicircles with radius 5 cm
= 10 × 10 – 2 × \(\frac{1}{2}\) × π × 5²
= 100 – 3.14 × 25
= 100 – 78.5 = 21.5 cm²
Similarly, Area of II + Area of IV = 21.5 cm²
So, area of the shaded region = Area of ABCD – Area of unshaded region
= 100 – 2 × 21.5 = 100 – 43 = 57 cm²

Question 5.
Find the area of the shaded region in figure, if ABCD is a square of side 7 cm. and APD and BPC are semicircles. (use π = \(\frac{22}{7}\))

Answer:
Given,
ABCD is a square of side 7 cm.
Area of the shaded region = Area of ABCD – Area of two semicircles with radius \(\frac{7}{2}\) = 3.5 cm
APD and BPC are semicircles.
= 7 × 7 – 2 × \(\frac{1}{2}\) × \(\frac{22}{7}\) × 3.5 × 3.5
= 49 – 38.5
= 10.5 cm²
∴ Area of shaded region = 10.5 cm

Question 6.
In figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm., find the area of the shaded region, (use π = \(\frac{22}{7}\)).
Answer:
Given, OACB is a quadrant of a Circle.
Radius = 3.5 cm; OD = 2 cm.
Area of the shaded region = Area of the sector – Area of △BOD

= 9.625 – 3.5 = 6.125 cm²
∴ Area of shaded region = 6.125 cm².

Question 7.
AB and CD are respectively arcs of two concentric circles of radii 21 cm. and 7 cm. with centre O (See figure). If ∠AOB = 30°, find the area of the shaded region. (use π = \(\frac{22}{7}\)).

Answer:
Given, AB and CD are the arcs of two concentric circles.
Radii of circles = 21 cm and 7 cm and ∠AOB = 30°
We know that,
Area of the sector = \(\frac{x}{360}\) × πr²
Area of the shaded region = Area of the OAB – Area of OCD

∴ Area of shaded region = 102.66 cm²

Question 8.
Calculate the area of the designed region in figure, common between the two quadrants of the circles of radius 10 cm each, {use π = 3.14)
Answer:
Mark two points P, Q on the either arcs.
Let BD be a diagonal of ABCD
Now the area of the segment

= 28.5 + 28.5 = 57 cm²

Side of the square = 10 cm
Area of the square = side × side
= 10 × 10 = 100 cm²
Area of two sectors with centres A and C and radius 10 cm.
= 2 × \(\frac{\pi r^{2}}{360}\) × x = 2 × \(\frac{x}{360}\) × \(\frac{22}{7}\) × 10 × 10
= \(\frac{1100}{7}\)
= 157.14 cm²
∴ Designed area is common to both the sectors,
∴ Area of design = Area of both sectors – Area of square
= 157 – 100 = 57 cm²
(or)
\(\frac{1100}{7}\) – 100 = \(\frac{1100-700}{7}\)
= \(\frac{400}{7}\)
= 57.1 cm²

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.2

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.2 Textbook Questions and Answers

Question 1.
Choose the correct answer and give justification for each.
(i) The angle between a tangent to a circle and the radius drawn at the point of contact is
a) 60°
b) 30°
c) 45°
d) 90°
Answer: [ d ]
If radius is not perpendicular to the tangent, the tangent must be a secant i.e., 90°.

(ii) From a point Q, the length of the tangent to a circle is 24 cm. and the distance of Q from the centre is 25 cm. The radius of the circle is
a) 7 cm
b) 12 cm
c) 15 cm
d) 24.5 cm
Answer: [ a ]

O – centre of the circle
OP – a circle radius = ?
OQ = 25 cm
PQ = 24 cm
OQ² = OP² + PQ²
[∵ hypotenuse² = Adj. side² + Opp. side²]
25² = OP² + 24²
OP² = 625 – 576
OP² = 49
OP = √49 = 7 cm.

iii) If AP and AQ are the two tangents a circle with centre O, so that ∠POQ = 110°. Then ∠PAQ is equal to

a) 60°
b) 70°
c) 80°
d) 90°
Answer: [ b ]
In □ OPAQ,
∠OPA = ∠OQA = 90°
∠POQ = 110°
∴ ∠O + ∠P + ∠A + ∠Q = 360°
⇒ 90° + 90° + 110° + ∠PAQ – 360°
⇒ ∠PAQ = 360° – 290° = 70°

iv) If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
a) 50°
b) 60°
c) 70°
d) 80°
Answer: [None]

If ∠APB = 80°
then ∠AOB = 180° – 80° = 100°
[∴ ∠A + ∠B = 90° + 90° = 180°]

v) In the figure XY and XV are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and XV at B then ∠AOB =
a) 80°
b) 100°
c) 90°
d) 60°

Answer: [ c ]

Question 2.
Two concentric circles of radii 5 cm and 3 cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle.
Answer:
Given: Two circles of radii 3 cm and 5 cm with common centre.

Let AB be a tangent to the inner/small circle and chord to the larger circle.
Let ‘P’ be the point of contact.
Construction: Join OP and OB.
In △OPB ;
∠OPB = 90°
[radius is perpendicular to the tangent]
OP = 3cm OB = 5 cm
Now, OB² = OP² + PB²
[hypotenuse² = Adj. side² + Opp. side², Pythagoras theorem]
5² = 3² + PB²
PB² = 25 – 9 = 16
∴ PB = √l6 = 4cm.
Now, AB = 2 × PB
[∵ The perpendicular drawn from the centre of the circle to a chord, bisects it]
AB = 2 × 4 = 8 cm.
∴ The length of the chord of the larger circle which touches the smaller circle is 8 cm.

Question 3.
Prove that the parallelogram circumscribing a circle is a rhombus.
Answer:

Given: A circle with centre ‘O’.
A parallelogram ABCD, circumscribing the given circle.
Let P, Q, R, S be the points of contact.
Required to prove: □ ABCD is a rhombus.
Proof: AP = AS …….. (1)
[∵ tangents drawn from an external point to a circle are equal]
BP = BQ ……. (2)
CR = CQ ……. (3)
DR = DS ……. (4)
Adding (1), (2), (3) and (4) we get
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + DC = AD + BC
AB + AB = AD + AD
[∵ Opposite sides of a parallelogram are equal]
2AB = 2AD
AB = AD
Hence, AB = CD and AD = BC [∵ Opposite sides of a parallelogram]
∴ AB = BC = CD = AD
Thus □ ABCD is a rhombus (Q.E.D.)

Question 4.
A triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D are of length 9 cm. and 3 cm. respectively (See below figure). Find the sides AB and AC.

Answer:
The given figure can also be drawn as

Given: Let △ABC be the given triangle circumscribing the given circle with centre ‘O’ and radius 3 cm.
i.e., the circle touches the sides BC, CA and AB at D, E, F respectively.
It is given that BD = 9 cm
CD = 3 cm

∵ Lengths of two tangents drawn from an external point to a circle are equal.
∴ BF = BD = 9 cm
CD = CE = 3 cm
AF = AE = x cm say
∴ The sides of die triangle are
12 cm, (9 + x) cm, (3 + x) cm
Perimeter = 2S = 12 + 9 + x + 3 + x
⇒ 2S = 24 + 2x
or S = 12 + x
S – a = 12 + x – 12 = x
S – b = 12 + x – 3 – x = 9
S – c = 12 + x – 9 – x = 3
∴ Area of the triangle

Squaring on both sides we get,
27 (x² + 12x) = (36 + 3x)²
27x² + 324x = 1296 + 9x² + 216x
⇒ 18x² + 108x- 1296 = 0
⇒ x² + 6x – 72 = 0
⇒ x² + 12x – 6x – 72 = 0
⇒ x (x + 12) – 6 (x + 12) = 0
⇒ (x – 6) (x + 12) = 0
⇒ x = 6 or – 12
But ‘x’ can’t be negative hence, x = 6
∴ AB = 9 + 6 = 15 cm
AC = 3 + 6 = 9 cm.

Question 5.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Verify by using Pythagoras Theorem.
Answer:
Steps of construction:

1. Draw a circle with centre ‘O’ and radius 6 cm.
2. Take a point P outside the circle such that OP =10 cm. Join OP.
3. Draw the perpendicular bisector to OP which bisects it at M.
4. Taking M as centre and PM or MO as radius draw a circle. Let the circle intersects the given circle at A and B.
5. Join P to A and B.
6. PA and PB are the required tan¬gents of lengths 8 cm each.

Proof: In △OAP
OA² + AP² = 6² + 8²
= 36 + 64 = 100
OP² = 10² = 100
∴ OA² + AP² = OP²
Hence AP is a tangent.
Similarly BP is a tangent.

Question 6.
Construct, a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Answer:
Steps of construction:

1. Draw two concentric circles with centre ‘O’ and radii 4 cm and 6 cm.
2. Take a point ‘P’ on larger circle and join O, P.
3. Draw the perpendicular bisector of OP which intersects it at M.
4. Taking M as centre and PM or MO as radius draw a circle which intersects smaller circle at Q.
5. Join PQ, which is a tangent to the smaller circle.

Question 7.
Draw a circle with the help of a bangle, take a point outside the circle. Con-struct the pair of tangents from this point to the circle measure them. Write conclusion.
Answer:
Steps of construction:

1. Draw a circle with the help of a bangle.
2. Draw two chords AB and AC. Perpendicular bisectors of AB and AC meets at ‘O’ which is the centre of the circle.
3. Taking an outside point P, join OP.
4. Let M be the midpoint of OP. Taking M as centre OM as radius, draw a circle which intersects the given circle at R and S. Join PR, PS which are the required tangents.

Conclusion: Tangents drawn from an external point to a circle are equal.

Question 8.
In a right triangle ABC, a circle with a side AB as diameter is drawn to intersect the hypotenuse AC in P. Prove that the tangent to the circle at P bisects the side BC.
Answer:
Let ABC be a right triangle right angled at P.
Consider a circle with diametere AB.
From the figure, the tangent to the circle at B meets BC in Q.
Now QB and QP are two tangents to the circle from the same point P.
QB = QP …….. (1)
Also, ∠QPC = ∠QCP
∴ PQ = QC (2)
From (1) and (2);
QB = QC Hence proved.

Question 9.
Draw a tangent to a given circle with center O from a point ‘R’ outside the circle. How many tangents can be drawn to the circle from that point? [Hint: The distance of two points to the point of contact is the same.
Answer:
Only two tangents can be drawn from a given point outside the circle.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.1

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.1 Textbook Questions and Answers

Question 1.
Fill in the blanks.
i) A tangent to a circle intersects it in ——— point(s). (one)
ii) A line intersecting a circle in two points is called a ———. (secant)
iii) The number of tangents drawn at the end of the diameter is ———. (two)
iv) The common point of a tangent to a circle and the circle is called ———. (point of contact)
v) We can draw ——— tangents to a given circle. (infinite)

Question 2.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Find length of PQ.
Answer:
Given: A circle with centre O and radius OP = 5 cm
\(\overline{\mathrm{PQ}}\) is a tangent and OQ = 12 cm
We know that ∠OPQ = 90°
Hence in △OPQ
OQ² = OP² + PQ²
[∵ hypotenuse² = Adj. side² + Opp. side²]
12² = 5² + PQ²
∴ PQ² = 144 – 25 .
PQ² = 119
PQ = √119

Question 3.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Answer:
Steps:

1. Draw a circle with some radius.
2. Draw a chord of the circle.
3. Draw a line parallel to the chord intersecting the circle at two distinct points.
4. This is secant of the circle (l).
5. Draw another line parallel to the chord, just touching the circle at one point (M). This is a tangent of the circle.

Question 4.
Calculate the length of tangent from a point 15 cm. away from the centre of a circle of radius 9 cm.
Answer:
Given: A circle with radius OP = 9 cm
A tangent PQ from a point Q at a distance of 15 cm from the centre, i.e., OQ =15 cm
Now in △POQ, ∠P = 90°
OP² + PQ² – OQ²
9² + PQ² = 15²
PQ² = 15² – 9²
PQ² = 144
∴ PQ = √144 = 12 cm.
Hence the length of the tangent =12 cm.

Question 5.
Prove that the tangents to a circle at the end points of a diameter are parallel.
Answer:
A circle with a diameter AB.
PQ is a tangent drawn at A and RS is a tangent drawn at B.
R.T.P: PQ || RS.
Proof: Let ‘O’ be the centre of the circle then OA is radius and PQ is a tangent.
∴ OA ⊥ PQ ……….(1)
[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]
Similarly, OB ⊥ RS ……….(2)
[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]
But, OA and OB are the parts of AB.
i.e., AB ⊥ PQ and AB ⊥ RS.
∴ PQ || RS.
O is the centre, PQ is a tangent drawn at A.
∠OAQ = 90°
Similarly, ∠OBS = 90°
∠OAQ + ∠OBS = 90° + 90° = 180°
∴ PQ || RS.
[∵ Sum of the consecutive interior angles is 180°, hence lines are parallel]

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.4

10th Class Maths 10th Lesson Mensuration Ex 10.4 Textbook Questions and Answers

Question 1.
A metallic sphere of radius 4.2 cm. is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Answer:
Given, sphere converted into cylinder.
Hence volume of the sphere = volume of the cylinder.
Sphere:
Radius, r = 4.2 cm
Volume V = \(\frac{4}{3}\)πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 4.2 × 4.2 × 4.2
= 4 × 22 × 0.2 × 4.2 × 4.2
= 4 x 22 x 0.2 x 4.2 x 4.2
= 310.464
Cylinder:
Radius, r = 6 cm
Height h = h say
Volume = πr²h
= \(\frac{22}{7}\) × 6 × 6 × h
= \(\frac{22 \times 36}{7} h\)
= \(\frac{792}{7} h\)
Hence, \(\frac{792}{7} h\) = 310.464
h = \(\frac{310.464 \times 7}{792}\) = 2.744cm
!! π can be cancelled on both sides i.e., sphere = cylinder

Question 2.
Three metallic spheres of radii 6 cm., 8 cm. and 10 cm. respectively are melted together to form a single solid sphere. Find the radius of the resulting sphere.
Answer:
Given : Radii of the three spheres r₁ = 6 cm r₂ = 8 cm r₃ = 10 cm
These three are melted to form a single sphere.
Let the radius of the resulting sphere be ‘r’.
Then volume of the resultant sphere = sum of the volumes of the three small spheres.


∴ 1728 = (2 × 2 × 3) × (2 × 2 × 3) × (2 × 2 × 3)
r³ = 12 × 12 × 12
r³ = 12³
∴ r = 12
Thus the radius of the resultant sphere = 12 cm

Question 3.
A 20 m deep well with diameter 7 m. is dug and the earth got by digging is evenly spread out to form a rectangu¬lar platform of base 22 m. × 14 m. Find the height of the platform.
Answer:
Volume of earth taken out = πr²h
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 20
= 770 m
Let height of plot form = H m.
∴ 22 × 14 × H = \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 20
H = \(\frac{35}{14}\) = \(\frac{5}{2}\) = \(2 \frac{1}{2} \mathrm{~m}\)
∴ The height of the plat form is \(2 \frac{1}{2} \mathrm{~m}\)

Question 4.
A well of diameter 14 m. is dug 15 m. deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 7 m to form an embankment. Find the height of the embankment
Answer:
Volume of the well = Volume of the embank
Volume of the cylinder = Volume of the embank
Cylinder :
Radius r = \(\frac{d}{2}\) = \(\frac{14}{2}\) = 7 cm
Height/depth, h = 15 m
Volume V = πr²h
= \(\frac{22}{7}\) × 7 × 7 × 15
= 22 × 7 × 15
= 2310 m³
Embank:

Let the height of the embank = h m
Inner radius ‘r’ = Radius of well = 7 m
Outer radius, R = inner radius + width
= 7m + 7m = 14 m
Area of the base of the embank = (Area of outer circle) – (Area of inner circle)
= πR² – πr²
= π(R² – r²)
= \(\frac{22}{7}\)\(\left(14^{2}-7^{2}\right)\)
= \(\frac{22}{7}\) × (14+7) × (14-7)
= \(\frac{22}{7}\) × 21 × 7
= 462 m²
∴ Volume of the embank = Base area × height
= 462 × h = 462 h m³
∴ 462 h m³ = 2310 m³
h = \(\frac{2310}{462}\) = 5 m.

Question 5.
A container shaped like a right circular cylinder having diameter 12 cm. and height 15 cm. is full of ice-cream. The ice-cream is to be filled into cones of height 12 cm. and diameter 6 cm., having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.
Answer:
Let the number of cones that can be filled with the ice-cream be ‘n’.
Then total volume of all the cones with a hemi spherical top = Volume of the ice-cream

Ice-cream cone = Cone + Hemisphere = πr²h

Cone:
Radius = \(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm
Height, h = 12 cm
Volume V = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 12
= \(\frac{22}{7}\) × 36
= \(\frac{792}{7}\)
Hemisphere:
Radius = \(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm
Volume V = \(\frac{2}{3}\)πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 3
= \(\frac{44 \times 9}{7}\)
= \(\frac{396}{7}\)
∴ Volume of each cone with ice-cream = \(\frac{792}{7}\) + \(\frac{396}{7}\) = \(\frac{1188}{7}\) cm³
Cylinder:
Radius = \(\frac{d}{2}\) = \(\frac{12}{2}\) = 6 cm
Height, h = 15 cm
Volume V = πr²h
= \(\frac{22}{7}\) × 6 × 6 × 15
= \(\frac{22 \times 36 \times 15}{7}\)
= \(\frac{11880}{7}\)
∴ \(\frac{11880}{7}\) = n × \(\frac{11880}{7}\)
⇒ n = \(\frac{11880}{7}\) × \(\frac{7}{1188}\) = 10
∴ n = 10.

Question 6.
How many silver coins, 1.75 cm in diameter and thickness 2 mm., need to be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Answer:
Let the number of silver coins needed to melt = n
Then total volume of n coins = volume of the cuboid
n × πr²h = lbh [∵ The shape of the coin is a cylinder and V = πr²h]

∴ 400 silver coins are needed.

Question 7.
A vessel is in the form of an inverted cone. Its height is 8 cm. and the radius of its top is 5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, 1/4 of the water flows out. Find the number of lead shots dropped into the vessel.
Answer:

Let the number of lead shots dropped = n
Then total volume of n lead shots = \(\frac{1}{4}\) volume of the conical vessel.
Lead shots:
Radius, r = 0.5 cm
Volume V = \(\frac{4}{3}\)πr³
= \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.5 × 0.5 × 0.5
Total volume of n – shots
= n × \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.125
Cone:
Radius, r = 5 cm;
Height, h = 8 cm
Volume, V = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 5 × 5 × 8
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 200

∴ Number of lead shots = 100.

Question 8.
A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter 4 \(\frac{d}{2}\) cm and height 3 cm. Find the number of cones so formed.
Answer:
Let the no. of small cones = n Then,
total volume of n cones = Volume of sphere Diameter = 28 cm.
Cones:
Radius r = \(\frac{d}{2}\)

Height, h = 3 cm

Total volume of n-cones = n . \(\frac{154}{9}\) cm³
Sphere:
Radius = \(\frac{d}{2}\) = \(\frac{28}{2}\) = 14 cm

No. of cones formed = 672.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.3

10th Class Maths 10th Lesson Mensuration Ex 10.3 Textbook Questions and Answers

Question 1.
An iron pillar consists of a Cylindrical portion of 2.8 m. height and 20 cm. in diameter and a cone of 42 cm. height surmounting it. Find the weight of the pillar if 1 cm³ of iron weighs 7.5 g.
Answer:
Volume of the iron pillar = Volume of the cylinder + Volume of the cone
Cylinder:

Radius = \(\frac{d}{2}\) = \(\frac{20}{2}\) = 10 cm
Height = 2.8 m = 280 cm
Volume = πr²h
= \(\frac{22}{7}\) × 10 × 10 × 280
= 88000 cm³
Cone:
Radius ‘r’ = \(\frac{d}{2}\) = \(\frac{20}{2}\) = 10 cm
height ‘h’ = 42 cm
Volume = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 10 × 10 × 42
= 4400 cm³
∴ Total volume = 88000 + 4400 = 92400 cm³
∴ Total weight of the pillar at a weight of 7.5 g per 1 cm³ = 92400 × 7.5
= 693000 gms
= \(\frac{693000}{1000}\) kg
= 693 kg.

Question 2.
A toy is made in the form of hemisphere surmounted by a right cone whose circular base is joined with the plane surface of the hemisphere. The radius of the base of the cone is 7 cm. and its volume is 3/2 of the hemisphere. Calculate the height of the cone and the surface area of the toy correct to 2 places of decimal.
(Take π = \(3 \frac{1}{7}\))
Answer:

Given r = 7 cm and
Volume of the cone = \(\frac{3}{2}\) volume of the hemisphere
\(\frac{1}{3}\)πr²h = \(\frac{3}{2}\) × \(\frac{2}{3}\) × πr³
∴ h = 3r
= 3 × 7 = 21 cm
Surface area of the toy = C.S.A. of the cone + C.S.A. of hemisphere
Cone:
Radius (r) = 7 cm
Height (h) = 21 cm
Slant height l = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{7^{2}+21^{2}}\)
= \(\sqrt{49+441}\)
= √490
= 22.135 cm.
∴ C.S.A. = πrl
= \(\frac{22}{7}\) × 7 × 22.135 = 486.990 cm²
Hemisphere:
Radius (r) = 7 cm
C.S.A. = 2πr²
= 2 × \(\frac{22}{7}\) × 7 × 7
= 308 cm²
C.S.A. of the toy = 486.990 + 308 = 794.990 cm²

Question 3.
Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 7 cm.
Answer:
Radius of the cone with the largest volume that can be cut out from a cube of edge 7 cm = \(\frac{7}{2}\) cm

Height of the cone = edge of the cube = 7 cm
∴ Volume of the cone V = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 7
= 89.83 cm³.

Question 4.
A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of right circular cone mounted on a hemisphere is immersed into the tub. The radius of the hemi¬sphere is 3.5 cm and height of cone outside the hemisphere is 5 cm. Find the volume of water left in the tub. (Take π = \(\frac{22}{7}\))
Answer:

The tub is in the shape of a cylinder, thus
Radius of the cylinder (r) = 5 cm
Length of the cylinder (h) = 9.8 cm
Volume of the cylinder (V) = πr²h
= \(\frac{22}{7}\) × 5 × 5 × 9.8
Volume of the tub = 770 cm³.
Radius of the hemisphere (r) = 3.5 cm
Volume of the hemisphere = \(\frac{2}{3}\)πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 3.5 × 3.5 × 3.5
= \(\frac{22 \times 12.25}{3}\)
= \(\frac{269.5}{3}\)
Radius of the cone (r) = 3.5 cm
Height of the Cone (h) = 5 cm
Volume of the cone V = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3.5 × 3.5 × 5
= \(\frac{192.5}{3}\)
Volume of the solid = Volume of the hemisphere + Volume of the cone
= \(\frac{269.5}{3}\) + \(\frac{192.5}{3}\) = \(\frac{462}{3}\) = 154 cm³
Now, when the solid is immersed in the tub, it replaces the water whose volume is equal to volume of the solid itself.
Thus the volume of the water replaced = 154 cm³.
∴ Volume of the water left in the tub = Volume of the tub – Volume of the solid = 770 – 154 = 616 cm³.

Question 5.
In the adjacent figure, the height of a solid cylinder is 10 cm and diameter 7 cm. Two equal conical holes of radius 3 cm and height 4 cm are cut off as shown in the figure. Find the volume of the remaining solid.

Answer:
Volume of the remaining solid = Volume of the given solid – Total volume of the two conical holes
Radius of the given cylinder (r) = \(\frac{d}{2}\) = \(\frac{7}{2}\) = 3.5 cm
Height of the cylinder (h) = 10 cm
Volume of the cylinder (V) = πr²h
= \(\frac{22}{7}\) × 3.5 × 3.5 × 10
= \(\frac{2695}{7}\)
= 385 cm³.
Radius of each conical hole, ‘r’ = 3 cm
Height of the conical hole, h = 4 cm
Volume of each conical hole,
V = \(\frac{1}{3}\)πr²h = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 4
= \(\frac{792}{21}\)
= \(\frac{264}{7}\)
Total volume of two conical holes = 2 × \(\frac{264}{7}\) = \(\frac{528}{7}\) cm³
Hence, the remaining volume of the solid

Question 6.
Spherical marbles of diameter 1.4 cm. are dropped into a cylindrical beaker of diameter 7 cm., which contains some water. Find the number of marbles that should be dropped into the beaker, so that water level rises by 5.6 cm.
Answer:
Rise in the water level is seen in cylindrical shape of Radius = Beaker radius
= \(\frac{d}{2}\) = \(\frac{7}{2}\) = 3.5 cm

Height ‘h’ of the rise = 5.6 cm.
∴ Volume of the ‘water rise’ = πr²h
= \(\frac{22}{7}\) × 3.5 × 3.5 × 5.6
= \(\frac{22 \times 12.25 \times 5.6}{7}\)
= 215.6
Volume of each marble dropped = \(\frac{4}{3}\)πr³
Where radius r = \(\frac{d}{2}\) = \(\frac{1.4}{2}\) = 0.7 cm
∴ V = \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.7 × 0.7 × 0.7
= 1.4373 cm³
∴ Volume of the ‘rise’ = Total volume of the marbles.
Let the number of marbles be ‘n’ then n × volume of each marble = volume of the rise.
n × 1.4373 = 215.6
= \(\frac{215.6}{1.4373}\)
∴ Number of marbles = 150.

Question 7.
A pen stand is made of wood in the shape of cuboid with three conical depressions to hold the pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Answer:
Volume of the wood in the pen stand = Volume of cuboid – Total volume of three depressions.
Length of the cuboid (l) = 15 cm
Breadth of the cuboid (b) = 10 cm
Height of the cuboid (h) = 3.5 cm
Volume of the cuboid (V) = lbh = 15 × 10 × 3.5 = 525 cm³.
Radius of each depression (r) = 0.5 cm
Height / depth (h) = 1.4 cm
Volume of each depressions V = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 0.5 × 0.5 × 1.4
= \(\frac{7.7}{3 \times 7}\) = \(\frac{1.1}{3}\) cm³
Total volume of the three depressions = 3 × \(\frac{1.1}{3}\)
= 1.1 cm³
∴ Volume of the wood = 525 – 1.1 = 523.9 cm³

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.2

10th Class Maths 10th Lesson Mensuration Ex 10.2 Textbook Questions and Answers

Question 1.
A toy is in the form of a cone mounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Determine the surface area of the toy. (Use π = 3.14)
Answer:

Diameter of the base of the cone d = 6 cm.
∴ Radius of the base of the cone
r = \(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm
Height of the cone = h = 4 cm
Slant height of the cone l = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{3^{2}+4^{2}}\)
= \(\sqrt{9+16}\)
= √25
= 5 cm
∴ C.S.A of the cone = πrl
= \(\frac{22}{7}\) × 3 × 5
= \(\frac{330}{7}\) cm²
Radius of the hemisphere = \(\frac{d}{2}\) = \(\frac{6}{2}\) = 3 cm
C.S.A. of the hemisphere = 2πr²
= 2 × \(\frac{22}{7}\) × 3 × 3
= \(\frac{396}{7}\)
Hence the surface area of the toy = C.S.A. of cone + C.S.A. of hemisphere
= \(\frac{330}{7}\) + \(\frac{396}{7}\)
= \(\frac{726}{7}\) ≃ 103.71 cm².

Question 2.
A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. The radius of the common base is 8 cm and the heights of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the total surface area of the solid. [Use π = 3.14]
Answer:
Total surface area = C.S.A. of the cone + C.S.A. of cylinder + C.S.A of the hemisphere.

Cone:
Radius (r) = 8 cm
Height (h) = 6 cm
Slant height l = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{8^{2}+6^{2}}\)
= \(\sqrt{64+36}\)
= √100
= 10 cm
C.S.A. = πrl
= \(\frac{22}{7}\) × 8 × 10
= \(\frac{1760}{7}\) cm²
Cylinder:
Radius (r) = 8 cm;
Height (h) = 10 cm
C.S.A. = 2πrh
= 2 × \(\frac{22}{7}\) × 8 × 10
= \(\frac{3520}{7}\) cm²
Hemisphere:
Radius (r) = 8 cm
C.S.A. = 2πr²
= 2 × \(\frac{22}{7}\) × 8 × 8
= \(\frac{2816}{7}\) cm²
∴ Total surface area of the given solid
= \(\frac{1760}{7}\) + \(\frac{3520}{7}\) + \(\frac{2816}{7}\)
T.S.A. = \(\frac{8096}{7}\) = 1156.57 cm².

Question 3.
A medicine capsule is ih the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the capsule is 14 mm. and the width is 5 mm. Find its surface area.
Answer:

Surface area of the capsule = C.S.A. of 2 hemispheres + C.S.A. of the cylinder
i) Now for Hemisphere:
Radius (r) = \(\frac{d}{2}\) = \(\frac{5}{2}\) = 2.5 mm
C.S.A of each hemisphere = 2πr²
C.S.A of two hemispheres
= 2 × 2πr² = 4πr²
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) × \(\frac{5}{2}\)
= \(\frac{550}{7}\)
= 78.57 mm².

ii) Now for Cylinder:
Length of capsule = AB =14 mm
Then height (length) cylinder part = 14 – 2(2.5)
h = 14 – 5 = 9 mm
Radius of cylinder part (r) = \(\frac{5}{2}\)
Now C.S.A of cylinder part = 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) × 9
= \(\frac{900}{7}\)
= 141.428 mm²
Now total surface area of capsule
= 78.57 + 141.43 = 220 mm²

Question 4.
Two cubes each of volume 64 cm³ are joined end to end together. Find the surface area of the resulting cuboid.
Answer:
Given, volume of the cube.
V = a³ = 64 cm³
∴ a³ = 4 × 4 × 4 = 4³ , Hence a = 4 cm
When two cubes are added, the length of cuboid = 2a = 2 × 4 = 8 cm,
breadth = a = 4 cm.
height = a = 4 cm is formed.

∴ T.S.A. of the cuboid
= 2 (lb + bh + lh)
= 2(8 × 4 + 4 × 4 + 8 × 4)
= 2(32 + 16 + 32)
= 2 × 80
= 160 cm²
∴ The surface area of resulting cuboid is 160 cm².

Question 5.
A storage tank consists of a circular cylinder with a hemisphere stuck on either end. If the external diameter of the cylinder be 1.4 m. and its length be 8 m. Find the cost of painting it on the outside at rate of Rs. 20 per m².
Answer:
Total surface area of the tank = 2 × C.S.A. of hemisphere + C.S.A. of cylinder.

Hemisphere:
Radius (r) = \(\frac{d}{2}\) = \(\frac{1.4}{2}\) = 0.7 m
C.S.A. of hemisphere = 2πr²
= 2 × \(\frac{22}{7}\) × 0.7 × 0.7
= 3.08 m².
2 × C.S.A. = 2 × 3.08 m² = 6.16 m²
Cylinder:
Radius (r) = \(\frac{d}{2}\) = \(\frac{1.4}{2}\) = 0.7 m
Height (h) = 8 m
C.S.A. of the cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × 0.7 × 8
= 35.2 m²
∴ Total surface area of the storage tank = 35.2 + 6.16 = 41.36 m²
Cost of painting its surface area @ Rs. 20 per sq.m, is
= 41.36 × 20 = Rs. 827.2.

Question 6.
A hemisphere is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the length of the cube. Determine the surface area of the remaining solid.
Answer:
Let the length of the edge of the cube = a units

T.S.A. of the given solid = 5 × Area of each surface + Area of hemisphere
Square surface:
Side = a units
Area = a² sq. units
5 × square surface = 5a² sq. units
Hemisphere:
Diameter = a units;
Radius = \(\frac{a}{2}\)
C.S.A. = 2πr²
= 2π\(\left(\frac{a}{2}\right)^{2}\)
= 2π\(\frac{a^{2}}{4}\) = \(\frac{\pi \mathrm{a}^{2}}{2}\) sq. units
Total surface area = 5a² + \(\frac{\pi \mathrm{a}^{2}}{2}\) = a²\(\left(5+\frac{\pi}{2}\right)\) sq. units.

Question 7.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm and its base radius is of 3.5 cm, find the total surface area of the article.

Answer:
Surface area of the given solid = C.S.A. of the cylinder + 2 × C.S.A. of hemisphere.
If we take base = radius
Cylinder:
Radius (r) = 3.5 cm
Height (h) = 10 cm
C.S.A. = 2πrh
= 2 × \(\frac{22}{7}\) × 3.5 × 10
= 220 cm²
Hemisphere:
Radius (r) = 3.5 cm
C.S.A. = 2πr²
= 2 × \(\frac{22}{7}\) × 3.5 × 3.5
= 77 cm²
2 × C.S.A. = 2 × 77 = 154 cm²
∴ T.S.A. = 220 + 154 = 374 cm².

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.1

10th Class Maths 10th Lesson Mensuration Ex 10.1 Textbook Questions and Answers

Question 1.
A joker’s cap is in the form of right circular cone whose base radius is 7 cm and height is 24 cm. Find the area of the sheet required to make 10 such caps.
Answer:

Radius of the cap (r) = 7 cm
Height of the cap (h) = 24 cm
Slant height of the cap (l) = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{7^{2}+24^{2}}\)
= \(\sqrt{49+576}\)
= √625
= 25
∴ l = 25 cm.
Lateral surface area of the cap = Cone = πrl
L.S.A. = \(\frac{22}{7}\) × 7 × 25 = 550 cm².
∴ Area of the sheet required for 10 caps = 10 x 550 = 5500 cm².

Question 2.
A sports company was ordered to prepare 100 paper cylinders without caps for shuttle cocks. The required dimensions of the cylinder are 35 cm length / height and its radius is 7 cm. Find the required area of thin paper sheet needed to make 100 cylinders.
Answer:

Radius of the cylinder, r = 7 cm
Height of the cylinder, h = 35 cm
T.S.A. of the cylinder with lids at both ends = 2πr(r+h)
= 2 × \(\frac{22}{7}\) × 7 × (7 + 35)
= 2 × \(\frac{22}{7}\) × 7 × 42 = 1848 cm².
Area of thin paper required for 100 cylinders = 100 × 1848
= 184800 cm²
= \(\frac{184800}{100 \times 100}\) m²
= 18.48 m².

Question 3.
Find the volume of right circular cone with radius 6 cm. and height 7 cm.
Answer:

Base radius of the cone (r) = 6 cm.
Height of the cone (h) = 7 cm
Volume of the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 7
= 264 c.c. (Cubic centimeters)
∴ Volume of the right circular cone = 264 c.c.

Question 4.
The lateral surface area of a cylinder is equal to the curved surface area of a cone. If their base be the same, find the ratio of the height of the cylinder to slant height of the cone.
Answer:
Base of cylinder and cone be the same.

CSA / LSA of cylinder = 2πrh
CSA of cone = πrl
The lateral surface area of a cylinder is equal to the curved surface area of cone.
∴ 2πrh = πrl
⇒ \(\frac{h}{l}=\frac{\pi r}{2 \pi r}\)
⇒ \(\frac{h}{l}\) = \(\frac{1}{2}\)
∴ h : l = 1 : 2

Question 5.
A self help group wants to manufacture joker’s caps (conical caps) of 3 cm radius and 4 cm height. If the available colour paper sheet is 1000 cm², then how many caps can be manufactured from that paper sheet?
Answer:

Radius of the cap (conical cap) (r) = 3 cm
Height of the cap (h) = 4 cm
Slant height l = \(\sqrt{r^{2}+h^{2}}\)
(by Pythagoras theorem)
= \(\sqrt{3^{2}+4^{2}}\)
= \(\sqrt{9+16}\)
= √25
= 5 cm
C.S.A. of the cap = πrl
= \(\frac{22}{7}\) × 3 × 5
≃ 47.14 cm²
Number of caps that can be made out of 1000 cm² = \(\frac{1000}{47.14}\) ≃ 21.27
∴ Number of caps = 21.

Question 6.
A cylinder and cone have bases of equal radii and are of equal heights. Show that their volumes are in the ratio of 3 : 1.
Answer:
Given dimensions are:
Cone:
Radius = r
Height = h
Volume (V) = \(\frac{1}{3}\)πr²h

Cylinder:
Radius = r
Height = h
Volume (V) = πr²h

Ratio of volumes of cylinder and cone = πr²h : \(\frac{1}{3}\)πr²h
= 1 : \(\frac{1}{3}\)
= 3 : 1
Hence, their volumes are in the ratio = 3 : 1.

Question 7.
A solid iron rod has cylindrical shape. Its height is 11 cm. and base diameter is 7 cm. Then find the total volume of 50 rods?
Answer:

Diameter of the cylinder (d) = 7 cm
Radius of the base (r) = \(\frac{7}{2}\) = 3.5 cm
Height of the cylinder (h) = 11 cm
Volume of the cylinder V = πr²h
= \(\frac{22}{7}\) × 3.5 × 3.5 × 11 = 423.5 cm³
∴ Total volume of 50 rods = 50 × 423.5 cm³ = 21175 cm³.

Question 8.
A heap of rice is in the form of a cone of diameter 12 m. and height 8 m. Find its volume? How much canvas cloth is required to cover the heap? (Use π = 3.14)
Answer:

Diameter of the heap (conical) (d) = 12 cm
∴ Radius = \(\frac{d}{2}\) = \(\frac{12}{2}\) = 6 cm
Height of the cone (h) = 8 m
Volume of the cone, V = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 8
= 301.71 m³.

Question 9.
The curved surface area of a cone is 4070 cm² and its diameter is 70 cm. What is its slant height?
Answer:
C.S.A. of a cone = πrl = 4070 cm²
Diameter of the cone (d) = 70 cm
Radius of the cone = r = \(\frac{d}{2}\) = \(\frac{70}{2}\) = 35 cm
Let its slant height be ‘l’.
By problem,
πrl = 4070 cm²
\(\frac{22}{7}\) × 35 × l = 4070
110 l = 4070
l = \(\frac{4070}{110}\) = 37 cm
∴ Its slant height = 37 cm.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.4

10th Class Maths 11th Lesson Trigonometry Ex 11.4 Textbook Questions and Answers

Question 1.
Evaluate the following:
i) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
Answer:
Given (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

ii) (sin θ + cos θ)² + (sin θ – cos θ)²
Answer:
Given (sin θ + cos θ)² + (sin θ – cos θ)²
= (sin² θ + cos² θ + 2 sin θ cos θ) + (sin² θ + cos² θ – 2 sin θ cos θ) [∵ (a + b)² = a² + b² + 2ab
(a – b)² = a² + b² – 2ab]
= 1 + 2 sin θ cos θ + 1 – 2 sin θ cos θ [∵ sin² θ + cos² θ = 1]
= 1 + 1
= 2

iii) (sec² θ – 1) (cosec² θ – 1)
Answer:
Given (sec² θ – 1) (cosec² θ – 1)
= tan² θ × cot² θ [∵ sec² θ – tan² θ = 1; cosec² θ – cot² θ = 1]
= tan² θ × \(\frac{1}{\tan ^{2} \theta}\) = 1

Question 2.
Show that (cosec θ – cot θ)² = \(\frac{1-\cos \theta}{1+\cos \theta}\)
Answer:

Question 3.
Show that \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A
Answer:
Given that L.H.S. = \(\sqrt{\frac{1+\sin A}{1-\sin A}}\)
Rationalise the denominator, rational factor of 1 – sin A is 1 + sin A.

[∵ (a + b)(a + b) = (a + b)²]; (a – b)(a + b) = a² — b²]
= \(\sqrt{\frac{(1+\sin A)^{2}}{\cos ^{2} A}}\)
= \(\frac{1+\sin A}{\cos A}\)
= \(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\)
= sec A + tan A = R.H.S.

Question 4.
Show that \(\frac{1-\tan ^{2} A}{\cot ^{2} A-1}\) = tan² A
Answer:

Question 5.
Show that \(\frac{1}{\cos \theta}\) – cos θ = tan θ – sin θ.
Answer:

Question 6.
Simplify sec A (1 – sin A) (sec A + tan A)
Answer:
L.H.S. = sec A (1 – sin A) (sec A + tan A)
= (sec A – sec A . sin A) (sec A + tan A)
= (sec A – \(\frac{1}{\cos A}\) . sin A) (sec A + tan A)
= (sec A – tan A) (sec A + tan A)
= sec² A – tan² A [∵ sec² A – tan² A = 1]
= 1

Question 7.
Prove that (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A
Answer:
L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
= (sin² A + cosec² A + 2 sin A . cosec A) + (cos² A – sec² A + 2 cos A . sec A) [∵ (a + b)² = a² + b² + 2ab]
= (sin² A + cos² A) + cosec² A + 2 sin A . \(\frac{1}{\sin A}\) + sec² A + 2 cos A . \(\frac{1}{\cos A}\)
[∵ \(\frac{1}{\sin A}\) = cosec A; \(\frac{1}{\cos A}\) = sec A]
= 1 +(1 + cot² A) + 2 + (1 + tan² A) + 2
[∵ sin² A + cos² A = 1; cosec² A = 1 + cot² A; sec² A = 1 + tan² A]
= 7 + tan² A + cot² A
= R.H.S.

Question 8.
Simplify (1 – cos θ) (1 + cos θ) (1 + cot² θ)
Answer:
Given that
(1 – cos θ) (1 + cos θ) (1 + cot² θ)
= (1 – cos² θ) (1 + cot² θ)
[∵ (a – b) (a + b) = a² – b²]
= sin² θ. cosec² θ [∵ 1 – cos² θ = sin² θ; 1 + cot² θ = cosec² θ]
= sin² θ . \(\frac{1}{\sin ^{2} \theta}\) [∵ cosec θ = \(\frac{1}{\sin \theta}\)]
= 1

Question 9.
If sec θ + tan θ = p, then what is the value of sec θ – tan θ?
Answer:
Given that sec θ + tan θ = p ,
We know that sec² θ – tan² θ = 1
sec² θ – tan² θ = (sec θ + tan θ) (sec θ – tan θ)
= p (sec θ – tan θ)
= 1 (from given)
⇒ sec θ – tan θ = \(\frac{1}{p}\)

Question 10.
If cosec θ + cot θ = k, then prove that cos θ = \(\frac{k^{2}-1}{k^{2}+1}\)
Answer:
Method-I:
Given that cosec θ + cot θ = k
R.H.S. = \(\frac{k^{2}-1}{k^{2}+1}\)

Method – II:
Given that cosec θ + cot θ = k ……..(1)
We know that cosec² θ – cot² θ = 1
⇒ (cosec θ + cot θ) (cosec θ – cot θ) = 1 [∵ a² – b² = (a -b)(a + b)]
⇒ k (cosec θ – cot θ) = 1
⇒ (cosec θ – cot θ) = \(\frac{1}{k}\)
By solving (1) and (2)

According to identity cos² θ + sin² θ = 1

Hence proved.